使用JavaScript从完整路径获取不带GET参数的文件名？

Question

Given /foo/bar/image.jpg?x=1&y=2 , how do I obtain image.jpg ?

https://stackoverflow.com/a/423385 provides a partial answer, but does not address the GET parameters.

Answer 1

You can use regex as others have suggested, but I find this more readable:

var src = '/foo/bar/image.jpg?x=1&y=2';
var img = src.split('/').pop().split('?')[0];
console.log(img);

Answer 2

Since the ? symbol separates the GET parameters from the URL, you can

str=str.split("?")[0]
filename = str.replace(/^.*[\/]/, '')

Answer 3

Try:

var str = "/foo/bar/image.jpg?x=1&y=2";
var fileName = str.split('/').slice(-1)[0].split('?')[0];

or, for a regex method:

var fileName = str.split('/').slice(-1)[0].match(/[^?]+/)[0];

Answer 4

You can use this regex:

/\/([^?\/]+(?=\?|$))/

and use captured grpup #1.

RegEx Demo

/        # matches a literal /
[^?\/]+  # matches 1 or more of any char that is not ? or /
(?=\?|$) # is a lookahead to assert that next position is ? or end of line